 internal stability of the system that means input output stable system does not mean that all other signal internal signal will be stable. So, let us consider one system that is we have a reference input the controller, then controller output is corrupted with a disturbance input disturbance, then plant output also part of with a output disturbance, then the measurement noise del y is d y is the measurement noise, output is corrupted with a noise or it is sometimes we call it is a sensor noise and this is feedback. Let us see here we have how many inputs are there these inputs are unwanted signal d u d y and d n d u is the input disturbance and d y is the output disturbance and d n is the measurement noise generally disturbances are same frequency as the reference or input low frequency signal, whereas in the measurement noises are high frequency signals. So, these things we keep in mind disturbance is a low frequency signal and measurements noises are the high frequency signals. So, let us call we have a if you see this one 4 inputs are there are d u d y d n out of 3 input signals are unwanted signal and we take the consideration each summer output is the which summer output that signal what we will get you will take as an output. So, we have a how many outputs are there this summer output e this summer output u of t u of p and this summer output is y and this summer output is y. Again we have a 4 outputs are there let us see what is the output of the system due to the effect of all inputs that is first term. So, our inputs are input signals are input signals are signals e then d u then d y then d n is the input signals then output signals are input signal are this is r sorry, this is r. So, output signal is e then summer each summer output is u p and this summer output is y this summer output is y n. So, if you see we have a 4 inputs 4 output system if we now you want to find out and further we consider that we have a single input single output system and that is linear time invariant system. We consider single input single output linear time invariant systems for this one we will see what is this output y of s form the Laplace transform output due to the all inputs. So, you first we will find out what is the Laplace transform of that means output y as a 4 components 1 component y 1 d 2 r s and y 2 component d 2 d u and y 3 component d 2 d y and y 4 component d 2 d n d n. So, total output having a 4 components each is due to the each inputs that r d u d y d n. So, let us find out first we consider since it is a linear time invariant system the output of y that means y 1 due to r 1 only keeping all inputs is assuming there is no inputs d u d y d y we want to find out the output y which is denoted by y 1 due to r. So, it is a simple our closed loop system standard closed loop system. So, loop we can write it g s c s forward path transformation divided by 1 plus loop transformation that is g s c s into r s. So, this component we can consider this component we can consider as if this is y 1 of s due to input of r s. Then we will find out what is the output contribution in this side due to this input only other inputs are assuming 0. So, if you see this one if you can read all this one you see this is the input is going that then our feedback is minus c s. So, it is a forward path transformation is g and loop transformation you see this is not there this is not there this is not there it is going here then this output then straight away with this controller it is feeding back to here. So, forward path is this one then loop transformation is g s c s into d capital d u d u of s plus transform of this one is d u of s. Next is we will find out what is this what is the output due to this input only. So, now see this is 0 this is 0 assume this is 0. So, now if I read all for this case I will read all and then I will explain how it is coming that 1 plus g s c s into d y of s read all this circuit when this input is only present other input is not there. So, I just read all this one d u. So, this is the output is going here this is the this is going from there this y I am feeding back feeding back. So, it is coming here c s g s this is not there this is not there this input. So, it is a c s g s and then it is feed back here minus. So, this is d y and this is your output y you can say this output is y 3 of t then what is this output here this one is 1 by this output this error signal you can say this nothing but a 1 by g s by s s 1 by 1 plus g s s s in this one this is a standard block diagram. So, you can find out what is the trans function between this and this forward path trans function is 1 and feed back trans function is g s s s s. So, that is we got it next part is left is next part what is left is plus I will write next part plus next page next part is plus. So, this will be due to this one due to this one what is the output here. Now, you can assume this signal you see what is the signal y n y is coming here d n is going here this then r is 0 d is 0 again this is 0. So, actually y plus d n is coming here and this is going here you one can think of it this d n is equivalent to r if this is coming and with minus. So, if I show it here with a minus sign here d n is the input and y is the feed back is coming from this one. So, it is same as our what is the trans function forward path trans function and loop trans function then g s c s divided by 1 plus g s c s minus d n s. So, this I can redraw if you like I can I will redraw 1 plus g s c s that will be minus d n of s. What I am telling it this signal I equivalently consider as if this signal is d n is here because this signal is y plus d n d n d n plus this will be minus y minus d n. So, minus d n I am taking into this one and then I find out the trans function between the y that is between d n to y or you redraw as you redraw. It is this one similar to this one you redraw this circuit when input is this one output is that one. So, it will come this one now this total output of the system due to the reference input due to the input disturbance due to the output disturbance and this is due to the noise in the measurements of sensor noise due to the sensor noise. So, one can see this one similarly this is for e s y s y s. So, this summer output we have a summer output here we have a summer output u of p here we have a summer output y n. Similarly, I can similar manner similar manner one can write e of s that I am not showing you can easily find out this one e s this e of s the signal e of small e is having a four components due to r 1 r due to d u due to d y due to d n. So, that you find out the trans function between input this and output this when these other the zeroes find out the trans function between the input this I agree when output is this one when these are zeroes and similarly you find out other inputs for other inputs also. So, if you calculate this one you will get 1 plus g s c s into r s minus g s 1 plus g s c s into d u of s minus d u of s minus 1 plus g s c of s d y of s plus 1 plus g s c s d n of s. Now, you see what is the error signal e of s in Laplace domain is contribution due to the four inputs that is what we are doing due to r s due to d u input disturbance due to d y that it out to output disturbance due to the measurement noise this similarly you can write there is an another output of the summer output is y y you have written. So, this e s is written then u of p I can write it u of p similarly u of p s is equal to c of s divided by 1 plus g of s c of s into r of s plus 1 plus g of s c of s d u of s minus c of s 1 plus g of s c of s into d y of s plus 1 plus this is g f plus this will be a next one is minus c of s 1 plus g of s into c of s into d n of s. You see thus you will get 1 plus g of s and you find out what is the transfer function between the input r and the signal y of s keeping all other input 0. It is just the way we have shown earlier the same way to find out only your input is r s that y p is the output again another contribution in y u of p due to the input u of s input and what is the y of p is the output was the transfer function between the input d u and u p find out the transfer function between d y u p find the transfer function of between input d from the input d n and u p. So, this transfer function you have obtained then another signal you have that is y n signal find out y of s transfer function between this reference input to the output this one from this input to the this output this input to this output this input to this output find out the transfer function you consider one at a time input into the systems. So, that you will get it if you do this one that you will get it y of n y of n s is equal to g of s into c of s divided by 1 plus g of s c of s into r of s this is due to the reference input. Then another term you will get g of s 1 plus g of s c of s into d u of s that is due to the input disturbance then another term 1 by 1 plus g of s c of s d y of s this is due to the y n due to the output disturbance then you have a 1 plus g s c s into d n this is due to the measurement noise input. So, we have a four inputs four outputs we got the that output expression into the Laplace transform four output expression in terms of Laplace transform due to the contribution of four inputs in Laplace transform. So, if you write it in more compact form in compact form form if you write it that means matrix and vector form if you write it then this will come we keep it right hand side is the output signals in Laplace transform. So, e of s then you have a u p of s then you have a y of s this order we have kept it you can put it any order that does not matter accordingly transform elements will change. So, what is our inputs inputs also r of s d u of s d y of s d n of s. So, now this four equation you this four equation you see that e s u p then your expression then y n expression then y y expression you if you write it in compact form you will get it one minus g of s minus one one then this is corresponding to that one then next is you will get it c then one minus c s minus c s then third one is your g c g of s c of s g of s one minus g of s into c of s then last expression last row is your g c g of s c of s g of s one minus g of s into c of s then last expression last row we can write g of s c of s g of s one and one and if you see g of s c of s g of c in all the expression e s all the one plus g g and c is common in all factors e s u p y p y and y n this factor is common. So, I am writing this factor one by this is one by that is is one plus g of s into c of s. So, this is one plus g of s into c of s denominated part is that one is common for all elements of this matrix. So, this is called transfer function matrix for multi input multi output case initially we as for single input single output case we used to call it is a transfer function for since it is a multi input multi we have a four inputs are there four outputs are there. So, naturally we have a this transfer function dimension is 4 cross 4 how many elements are there 16 element. So, this feedback system is internally stable that this feedback system this is the what in general it is nothing but a if you write it that output that all input signal is mapped to a output signal through a what is called a matrix which we call transfer function matrix. So, our conclusion is the feedback system which is described into transfer function model which is described in transfer function is internally stable if and only if necessary and sufficient condition each element that one by one plus g c c g minus g of s divided by one plus g s c s must be each element of this matrix must be stable. In the other sense the all the poles must be left up of the s plane then the system is internally stable means this indicates all internal signals e u p then y n then y are all stable then we will call the overall system is stable in the in that sense it is not that only input output point of view we told the system is input output means only r s and y s. So, the system is internally stable if and only if each element of the transfer function each element of the transfer function matrix has stable poles or Harwit or Harwit this is the internal stability of the system one has to judge like this way. Let us see the effect of this how we can reduce the disturbance effect at the output how we can reduce the noise effect at the output. Because consider now recall the our output equation recall the output equation and output equation is we have written the first this will write it this equation first this equation. So, if you write it this one y of s is equal to g of s c of s one plus g of s is equal to g of s c of s into r of s this is due to the input and next is g of s one plus g of s c of s into d of u of s then minus or plus you can write it one plus g of s c of s into d y of s minus g of s c of s one plus g s c of s d n of s. Again this is the output of the system what we have considered is affected by reference input input disturbance output disturbance and measurement noise or sensor noise. Then our next job will be how we can minimize the effect of disturbance and noise at the output that is our next target that how we will design a controller. So, that effect of disturbance at the output will be minimized. So, let us rewrite this equation we can rewrite that equation that we can write one by let us call one by g of s is defined by s 0 of s then this will be a at 0 of s and this we denoted by t 0 of s this transformation. So, we can write it t 0 of s into r of s then g of s into s 0 of s into d u of s plus s 0 of s d y of s minus t 0 of s into d u of s d n of s. Now, we define that this s 0 of s which is nothing but a one plus g s c s we call this is a sensitivity function sensitivity b t sensitivity b t function it has implication on the while we will design the system or when we will design the controller the sensitivity sensitivity function. Then t 0 of s is called t 0 of s is a g of s c of s one plus g of s c of s is called complementary sensory sensitivity function c t b t function. So, and it can be noted that you can note that t 0 of s plus s 0 of s that value is one. So, t 0 of s is now we will see look at this expression that I want to I want the effect of output disturbance at the input should be as small as possible. Now, what we can do it as you can remember that I mentioned that this signal input disturbance and measurement signal this signal are low frequency signal. Whereas, measurement noise is high frequency signal this is high frequency high frequency signal and these are low frequency signal low frequency signal this and this because it disturbance is a low frequency signal. So, look at this expression since it is a low frequency signal and our plant is most strictly proper trans function most of the particular plant is a strictly proper trans function and it is a plant is a low pass filter. So, naturally if you see the frequency response of g the magnitude will decrease with increasing in frequency the high frequency the magnitude of g will be a small and with a high frequency with increasing in frequency it will magnitude will decrease. And now our job job is if the we can make it the sensitivity function at low frequency is very small then effect of disturbance at the output due to the output disturbance will be small then how you make it this s is small at low frequency. This it is obvious that g s c of s you have to design in such a way that product of g s and c h which is called loop trans function must be very high at low frequency. If it is a very high it magnitude at low frequency then 1 by 1 plus g s will be small because this quantity is large. So, the effect of disturbance will be at the output will be as small as possible by selecting the this is called the loop trans function loop trans function. The loop trans function value at low frequency is very high this we can make it and if you can make it this one this effect of disturbance at the output will be small. And here also you can see this is the low frequency signal this now this things you see this 1 by g of s is small and g s is the strictly proper trans function the effect of d u also will be small at the output. But, this term will not be affected because this quantity you see this g h it is it is it is affected with a high frequency signal not low frequency signal. So, this picture will not come when it is a low frequency signals are accepted with the low frequency signals. And this magnitude is small for this one when it is accepted with a noise this term will not come to the picture because d n is the high frequency signal we know this high frequency is the magnitude the loop trans function value is very small. So, this is small 1 plus this small. So, it is nothing but a this is small compared to 1 you can neglect it. So, g s c s is very small when it is excited with a high frequency signal. So, we can now summarize this all these things into a this thing design our design of objective you recall our equation y of s is equal to t 0 of s r s plus g h 0 of s d u of s plus h 0 of s d y of s minus t 0 of s d n of s. So, from this equation we can say for one design objective for good output disturbance of the rejection what we have to do it we have to do it output disturbance that output disturbance this term that s 0 must be small in order to make s 0 the loop transform g s c s should be high at low frequency. So, in order to this s of s must be or s of j omega must be mod of s of j omega must be very less than 1. This can be achieved if the loop trans function loop trans function that mod g of j omega h c c of j omega is very greater than 1. So, this is one if we can make it s is equal to 0 then you see the effect of disturbance and output will not come into the picture. If s of j omega mod of this one is make it will 0 again at low frequency region then perfect output disturbance rejection is possible. This is one condition the output disturbance next is good for good tracking the reference input we have output should track the reference input. If output should track the reference input then t 0 of s must be equal to 1. I told you r s d u d y all are low frequency signals. So, next objective design objective even if you want to achieve next stage for good set point tracking what does it mean the output should track the reference input that is this one the output should track the reference input in order to achieve this one. If you see this equation if you this t 0 of s the small because r s is a small frequency. So, signal small frequency signal. So, the t s in the small frequency range t 0 of s must be equal to 1. So, t 0 of s nearly equal to 1 this can be achieved if loop trans function g of j omega mod c of j omega mod is very very greater than 1. Now, if it is what is t 0 of s t 0 of s c g c s divided by 1 plus t 0 expression if you see c s g s c of s 1 plus g s c of s agree. S is equal to j omega this quantity if you want to make it very very greater than 1 very very greater than 1. Then what is this quantity will begin what will get it 1 is neglected with this one. So, this can say it will be 1 if you can loop trans function if it is high in low frequency and that is possible with since the plant is stable and is a strictly proper trans function that low frequency region the value of g is high and also design a controller c of s in that way it is a strictly proper trans function and also the magnitude of c j at low frequency is very high the product will be very high at low frequency. So, third objective of the design is for good noise suppression for good noise suppression. Now, see that this noise suppression if you want to make it this one then this should be a 0 agree, but mind it this trans function what is this one this is excited with a high frequency signal this is a high frequency signal signal. So, for suppression of high frequency signal that suppression of the noise at the output because noise is the high frequency signal this must be equal to 0. You have to make it 0 then you have to make that loop trans function at high frequency the magnitude of that one must be nearly equal to 0 or as small as possible. So, for good suppression that t 0 of s must be 0. As small as possible if t 0 of s nearly equal to 0 then we would be then would be perfect perfect noise suppression then there would be perfect noise suppression. So, how to say you know I just mentioned it g of j omega at high frequency the magnitude is small similarly c of j omega at high frequency the magnitude is product of this one will be very small. So, if you can design a controller in such a way at high frequency the magnitude of that g of plant magnitude is nearly equal to very small and g of s does not increase with change in high frequency range because it is decreasing with high frequency range then product is small. So, we can write it note at high frequency g of j omega magnitude will be small because of it is a strictly proper trans function and stable one small since g of j is strictly proper trans function this is one way and next since c of j omega c of j omega does not increase does not increase with frequency does not increase with frequency with high frequency. So, the t 0 of j omega mod can be made can be made small at high frequency range at high frequency range high frequency range so, this the third objective is good noise measurement 4th one is your for less control energy to be supplied to the systems control energy afford for less control afford energy afford. So, this can be achieved by c of s a 0 of s must be c of s must be c of s as small as possible as possible. Now, look at this one that expression the control afford expression I will just show you the control afford expression this one. Now, this control afford afford u of t is influenced by r of s d of s y of s d of s d of s d of s d on f s. Now, see this one this c s divided by 1 plus g s of s. So, c s into s 0 of s so the control afford will be u of t or afford will be a small if you can make small because it is a low frequency range this. So, if you can make it so this is also you can make it small at low frequency because loop trans function is high loop trans function here is high. So, this is possible provided you can make it this as small as possible because this quantity at c s and s 0 you may make it because I can make it this is small because the loop trans function at low frequency our designer choice is that should be a loop trans function will be high which in time s 0 should be low. So, this way we can reduce the control afford in the into the system by c 0 of s by making c 0 of s as small as possible. So, our desirable shape of the loop trans function shape of loop trans function like this way this way this is omega and let us call this is our we are plotting loop trans function g of s c of j omega mod in this directions agree and we have a this region we are showing and this is the region. So, frequency response of our loop trans function should be like this way. So, this indicates the gain above this level in the low frequency. That means loop gain should be more than loop gain this is the loop gain loop gain should be more than the this level more than this level at low frequency. This is the loop gain with loop gain frequency versus magnitude of the magnitude of the loop trans functions and loop trans function gain magnitude of the loop trans function must be greater than this level. Whereas, the loop gain at high frequency this is at low frequency whereas, the loop gain at high frequency must be less than this level must be less than this level. So, this is the you can say this is the gain loop gain below this level in high frequencies the loop gain you can write it loop gain this is loop gain loop gain means g s c s this is the distribution of desirable shape of our loop gain. Then we will be able to achieve the four other design objectives that good tracking good disturbance rejection good noise suppression than less control effort. If you can make the desirable shape of the loop trans function like this way that means at low frequency the loop gain must be more than this level and the loop gain at high frequency must be less than this level that is what you can do and this point is called the mid frequency mid frequency and this determines bandwidth and stability margin. So, we have seen whether our design objective is there in true sense if you see in order to when we are going to design the what is called our controller to take care of these all design objectives. Our plant model we must know, but generally when we have given the actual plant and the plant model when we are doing that some of the issues we are not taken into consideration when we are modeling the plant. So, the model what we will get it that is called the approximate mathematical model of the plant. So, there are plant uncertainties is there depending upon the two types of uncertainties structure uncertainties and the unstructured uncertainties. Suppose for example, if you have a simple resistance inductance and capacitance connected in series and that circuit is accepted with a input signal and if you find out the trans function between the input signal and the output is the current flowing to the circuit then trans function will contain the what is called numerator and denominator polynomial in S. So, in second order system you will get the denominator polynomial of order 2 and each coefficient of the polynomial is nothing but a system parameters system parameter means in this case it is a resistance inductance and capacitance. So, this resistance and inductance and capacitance will change with time today we have found out the trans function for the value of R L and C may be three months after the parameter value may change it. So, it is model parameter it may be we will change it, but we have designed the controller based on the nominal model then after three months the plant model will change. So, same controller whether it will be stable stabilize the plant or not that is the issue of our study next. So, next is your plant uncertainty plant uncertainty plant uncertain arises arises from the inevitable of the discrepancy between the plant and the model between the true plant and its model. So, plant uncertainty is there. So, there are two types of uncertainty and two types of uncertainty structured uncertainty two is unstructured uncertainty. Structure uncertainty I just mentioned it that suppose we have model the system of R L C circuit now after may be one year or six months the parameters of R L C may change it again and we know particularly which coefficient of the denominator of the trans function polynomial is change that we can know that you know exactly the position where the parameters are changed which coefficient parameters are changed. So, that is will be called structure uncertainty and unstructured we have do not have much knowledge about the model uncertainties about that one. So, our problem is now robust before that I will check what do you mean by we will just discuss that unstructured uncertainty. Suppose first we consider the case additive uncertainties additive uncertainties. We have a system we have a controller we have a plant and this is the model of the true plant this one when we are doing the plant model some of the issues we have not taken into considerations either if you get the model from frequency response some of the high frequency effect high frequency signal effect where the plant magnitude is very small and it is highly corrupted with the noise that part is we are not able to model that part high frequency part of the model system will not be able to model it. So, this is the plant model and we consider whatever the discrepancy between the two plant and the model is denoted by a delta a of delta delta a of s this is the and this output then it is a feedback r s this is y s. So, this is the plant model and this is and there is a discrepancy between the true plant and the model and its model and that discrepancy is compensated with a delta a of s. We do not know about the delta a that perturbation only we know its frequency response if you see what is the upper bound of that perturbations we know it. So, and we made an assumption this perturbation we made it is stable transfunction that perturbation stable transfunction. Now, if you see this one what is this model is that that one what is this model it is nothing but a g m of s plus delta a of s this block g m delta a of s. So, our problem is to study the stability of that one. So, let us call this is u this is b. So, let us call the controller c of s is designed based on the model parameters that model parameters g m of s model transformation now our question is whether the how much this controller can withstand the perturbation before the closed loop system becomes unstable. So, I have to find out the bound of this already perturbations what is the maximum perturbation bound that controller can withstand before going to unstable. So, then we have to how we have to design that one that is our problem. So, next class we will see that how to handle such type of problems to design a controller.